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Popular >

4y^'+y=0

Solution

4 y^{^{'}} + y = 0

Solution

y = e^{\frac{- t + c _{1}}{4}}

Solution steps

4 y^{'} (t) + y = 0

Solve separable ODE:

y = e^{\frac{- t + c _{1}}{4}}

y = e^{\frac{- t + c _{1}}{4}}

Popular Examples

tangent of f(x)=x^2+8,(-2,12)tangent of

f (x) = x^{2} + 8, (- 2, 12)

d/(d{y)}(-({y})/(({x)^2+{y}^2+{z}^2)^{3/2}})

\frac{d}{d y} (- \frac{y}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}})

integral of (3x^2-x)

\int (3 x^{2} - x) d x

derivative of e^{3-2x}

\frac{d}{d x} (e^{3 - 2 x})

integral of (x^2+1)^9

\int (x^{2} + 1)^{9} d x

Frequently Asked Questions (FAQ)

What is the solution for 4y^'+y=0 ?
The solution for 4y^'+y=0 is y=e^{(-t+c_{1})/4}